study material kinematics for aieee

7/29/2019 Study Material Kinematics for AIEEE

1/60

KINEMATICS

SYLLABUS :Motion in one and two dimension, Projectile motion, Relative Velocity & Circular

Motion.

POSITION VECTOR

If the coordinates of a particle are given by (x2, y2, z2) its position vector with

respect to (x1,y1,z1) is given byr = (x2 - x1) i + (y2 - y1) j +(z2 - z1) k. Usually,

position vector with respect to the origin (0,0,0) is specified and is given byr =x i +y j +z k

DISPLACEMENT

Displacement is a vector quantity. It is the shortest distance

between the final and initial positions of a particle. If 1r

is the initial position vector and 2r is the final position

vector, the displacement vector is given by 12

= rrr .

The magnitude of the displacement is given by

( ) ( ) ( )2122

122

12 zzyyxx ++

This is nothing but the straight line distance between two points (x1,y1,z1) and(x2,y2,z2). The displacement is independent of the path taken by the particle inmoving from (x1 , y1, z1) to (x2, y2, z2)

DISTANCE :

If a particle moves along a curve, the actual

length of the path is the distance. Distance isalways more than or equal to displacement.

Illustration 1 :

A car travels along a circular path of radius (50 /)m with a speed of 10 m/s. Findits displacement and distance after 17.5 sec.

Solution :Distance = (speed) time = 10 (17.5) = 175 mPerimeter of the circular path = 2 (50/) = 100 m

The car covers 143 rounds of the path

)z,y,x(P 111)z,y,x(Q 222

A

B

1

r

r

2

r


2/60

If the car starts from A, it reaches B and the displacement is the shortestdistance between A and B

Displacement = 22 RR + =

=250

2 R m.

INSTANTANEOUS AND AVERAGE VELOCITY :

Ifr is the displacement of the particle in time t, the average velocity is

given by

V average =

t

r

tt

rr

=

12

12

= Find value - Initial value. The above definition is valid for any magnitude

of large or small. But when is infinitesimally small, the instantaneous

velocity is obtained.

V instantaneous = Ltt 0 t

r

=dt

rd

In normal notation, velocity refers to the instantaneous velocity.

SPEED :

Speed =time

cetanDis

When the time under consideration is very small, distance becomes equal tothe displacement and speed becomes the magnitude of instantaneous velocity.speed is represented only by its magnitude where as velocity is represented bymagnitude as well as direction.

INSTANTANEOUS AND AVERAGE ACCELERATION :

If

V is the change in velocity in time t, average acceleration is given by

averagea =

t

V

.

When becomes infinitesimally small,t

VLtt

0 = dt

Vd

which gives the instantaneous acceleration.

In normal notation, acceleration refers to the instantaneous acceleration.

==

dt

rd

dt

d

dt

Vda =

2

2

dt

rd


3/60

It may be noted here that magnitude of2

2

dt

rd

is not equal to2

2

dt

rdalways (as in

the case of circular motion)

Illustration 2 :

A bus shuttles between two places connected by a straight road with uniform speedof 36 kmph. If it stops at each place for 15 minutes and the distance between the twoplaces is 60 km, find the average values of (a) Speed (b)Velocity

(c) acceleration between t = 0 and t = 2 hours and the instantaneous values of

(d) Velocity (e) acceleration at t = 2 hrs.

Solution :

Time taken for forward trip = 36

60= 3

5hrs.

Time of stoppage = 15 min = 0.25 hrs.

Time available for return trip = 2 - 5/3 - 0.25 = 1/12 hrs.

Distance travelled in the return trip = (36) 1/12 = 3 km.

a) Average speed =timeTotal

distanceTotal =2

360 + = 31.5 kmph.

b) Average velocity =Time

ntDisplaceme=

2

360 = 28.5 kmph

c) Average acceleration =Time

velocityinchange

=t

VV

12 =( ) ( )

2

3636 += - 36 km/H2 = -

360

1m/s2

d) Velocity at t = 2 hours = - 36 kmphe) Acceleration at t = 2hours = 0 as there is no change in velocity

Illustration 3 :

A car travels towards North for 10 minutes with a velocity of 60 Kmph, turnstowards East and travels for 15 minutes with a velocity of 80 kmph and then turnstowards North East and travels for 5 minutes with a velocity of 60 kmph. For thetotal trip, find a) distance travelled b) displacement c) average speed d)average velocity and e) average acceleration.

Solution :Total time taken = (10 + 15 + 5)min = 1/2 hour

kmd 31 =

kmd 60=


4/60

a) Distance travelled = d1 + d2 + d3

= 60

60

10+ 80

60

15+ 60

60

5

= 10 + 20 + 5 = 35 km

b) displacement

++= 321 SSSS

= 10 j + 20 i +5 cos 450 i + 5 sin 45 j

= 23.5 i + 13.5 j

Magnitude of displacement = ( ) ( )22 513523 .. + ~ 27 km

c) Average speed =

hr

kmdiTotal

=

2

1

35

Time

travelledstance= 70 kmph.

d) Average velocity =Time

ntDisplaceme = ( )jij.i. 2747

2

1513523 +=+ kmph.

at an angle with the East given by Tan =523

513

.

.

Magnitude of average velocity = 22 2747 + = 54 kmph

e) Average acceleration =Time

yin velocitchange

= Time

velocityInitialvelocityFinal

=( )

2

6045604560 0 jjsinicos +

= ( )ji 921 km/H2

at an angle with the East given by Tan =21

9

Magnitude of average acceleration = 22 921 + ~ 23 km/H2 ~ 1.8 x 10-

3 m/s2

Illustration 4 :

A car moving along a circular path of radius R with uniform speed covers an angle during a given time. Find its average velocity and average acceleration during thistime.

Solution :

Let V be the speed of the car

V =time

Distance=

t

Rwhere is in radians.

Displacement= + cosRRR 222 2 from the triangle OAB

A V

B

Vo

1S

2

S S

3

S


5/60

= 2R sin /2

Average velocity =Time

Diplacemnt=

V

R

sinR2

2=

22 sinV

Average acceleration =t

V

time

velocityinChange =

V = + cosVVV 222 2 = 2 V sin2

Average acceleration =

V

R

sinV2

2=

R

sinV2

2 2

When is small sin ~ and

Average velocity =

22 sinV

= V2

V2

=

Average velocity = Instantaneous velocity for small angulardisplacements

Average acceleration =

=

R

V

R

sinV2

22

222

=R

V2

Average acceleration = Instantaneous acceleration for small angulardisplacements.

KINEMATICAL EQUATIONS : ( CONSTANT ACCELERATION ) :

V = u + atV2 - u2 = 2aSS= ut + 1/2 at2The above equations are valid only for constant acceleration and in aparticular direction. u,v and s must be taken with proper sign. Usually the

direction of u is taken as positive and the sign of other variables are decidedwith respect to this direction.Displacement during the nth second = Sn - Sn-1

= u +2

a(2n - 1)

It may be noted here that this is not the distance travelled in the n th second.

Illustration 5 :A particle is vertically projected upwards with an initial velocity of 22.5 m/s.Taking g=10 m/s2 find a) velocity b) displacement c) distance travelled in t = 4 sec

and d) displacement and distance travelled in 3rd second

V

V

V


6/60

Solution :Taking the upward direction positive

a) V = u + (-g) t = 22.5 - 10 (4) = -17.5 m/s 17.5 m/s down wardsb) S = ut + 1/2 (-g) t2 = 22.5 (4) - 1/2 (10) 42 = 10 m

c) Time to reach the top most point = t0 and at the top mostpoint velocity becomes zero.

V = u - gt0 0 = 22.5 - 10 (t0) t0 = 2.25 secDistance travelled in 4 sec = d1 + d2

d1 = u t0 - 1/2 g20t = 22.5 (2.25) - 1/2 (10) (2.25)2 = 25.3 m

d1 can be found from V2 - u2 = 2a S also.

0 - (22.5)2 = 2(-10) d1 d1 =( )

20

522 2.= 25.3 m

d2 can be found from S = ut + 1/2 at2 applied along the down ward

direction starting from the top most pointd2 = 0 (t - t0) + 1/2 g (t - t0)2 = 1/2 (10) (4 - 2.25)2 = 15.3 m

Distance travelled in 4 sec = 25.3 + 15.3 = 40.6 mDisplacement in 4 sec = d1 - d2 = 25.3 - 15.3 = 10 mDisplacement can also be found directly by applying S = ut +1/2 at2 along the verticalDisplacement in 4 sec = 22.5 (4) - 1/2 (10) (4)2 = 10M

d) 3rd second is from t = 2 sec to t = 3 sec.

Displacement in the 3rd second = u +2

a(2n - 1)

= 22.5 -210 (6 - 1) = -2.5 m

When there is no change in the direction of the motion along a straight line,distance will be equal to displacement. When the particle reverses itsdirection during the time under consideration, distance will be more than thedisplacement and the time at which the reversal is taking place must befound.

When the particle reverses its direction, its velocity becomes zero.

using V = u + at, 0 = 22.5 - 10 (t0) t0 = 2.25 sec

d = d1 + d2using the formula S = ut + 1/2 at2

d1 = [22.5 (2.25) - 1/2 (10) (2.25)2 ] - [22.5 (2) - 1/2 (10) (2)2]

= 0.31 m

Along the downwards vertical starting from the top

d2 = 0 (3 - 2.25) + 1/2 (10) (3 - 2.25)2 = 2.81 m

d = 0.31 + 2.81 = 3 .12 m

KINEMATICAL EQUATIONS ( VARIABLE ACCELERATION ) :

When the acceleration is variable, the kinematical equation take the form

sect 3=sect 2=

sec.t 252=

1d 2d

1d

2d


7/60

V =dt

dxa =

2

2

dt

xd

dt

dV=

a = dx

VdV

dt

dx

dx

dV

=

V = u +

t

dta0

x = ut + dtdta

tt

00

and V2 - u2 = 2 x

dxa

0

Illustration 6 :

The position coordinate of a particle moving along a straight line is given by x = 4 t3-3t2+4t+5. Find a) Velocity and acceleration as a function of time b) Displacementas a function of time c) the time at which velocity becomes zero and the acceleration

at this time d) the time at which acceleration becomes zero and the velocity at thistime.

Solution :

a) V =dt

dS=

( )

dt

xxd 0 =dt

dxWhere x0 is the initial position coordinate

which is a constant

=dt

d(4t3 - 3t2 + 4t + 5) = 12 t2 - 6t + 4

a =dt

dv=

dt

d(12t 2 - 6t + 4) = 24t - 6

b) Displacement = (position coordinate at time t) - (position coordinate att = 0)

= (4t3 - 3t2 + 4t + 5) - (5)

= 4t3 - 3t2 + 4t

c) When V = 0, 12 t2 - 6t + 4 = 0 t =12

4893

since this value is imaginary, the velocity never becomes zero.

d) When a = 0, 24t 6 =0 and t = 24

6= 4

1units and the velocity of the

particle at this time, V= 124

134

4

16

4

12

=+

units

Illustration 7 :

The velocity of a particle moving in the positive direction of the x axis varies as V =

x where is a positive constant. Assuming that at the moment t = 0 the particlewas located at the point x = 0, find a) the time dependence of the velocity and theacceleration of the particle b) the mean velocity of the particle averaged over thetime that the particle takes to cover the first S meters of the path.


8/60

Solution :

a) V =

dt

dx= x =

tx

dt

x

dx

00

2 x = t and x =4

22t

V =2

2t

and a =dt

dV=

2

2

b) Mean velocity =time

ntDisplaceme

Displacement = S, and the time taken for this displacement t =

S2

Mean velocity =( )

S

S

2=

2

S

Mean velocity can also be found from the following formulae

V mean =

dx

dxVwhen V is a function of x

and V mean =

dt

dtV

when V is a function of time

KINEMATICAL EQUATIONS IN VECTOR FORM ( CONSTANT ACCELERATION )

tauV

+=

= S.au.uV.V 2

2

21 tatuS +=

The above equations are useful in 2 and 3 dimensional motion.

Illustration 8 :A particle moving on a horizontal plane has velocity and acceleration as shown inthe diagram at time t = 0. Find the velocity and displacement at time 't'.


9/60

Solution :METHOD - Iu = u cos300 i + u cos 60 j =

2

3u i + j

u

2

a = - a cos 45 i - a cos45 j =

2

a i -

2

aj

V = tau

+ = it

au

22

3+ jt

au

22

The magnitude of the velocity =22

2222

3

+

t

auatu

2

2

1tatuS

+= = jt

at

uit

aut

+

22

22

1

222

1

2

3= Sx i + Sy

j

The magnitude of the displacement = 22 ySSx +

METHOD - II

This can be solved by vector addition method also. It may be noted

here thatu t will be along the direction of

u , ta

and 2

2

1ta

will be

along the direction of

a

tauV

+=

Since the angle betweenu and ta

is 1650, the magnitude of the

velocity is

( ) ( ) 022 1652 cosatuatu ++

2

2

1tatuS

+=

Since the angle between

u t and 221ta

is 1650, the magnitude of the

displacement is ( ) ( ) 022

22 1652

12

2

1cosatutatut

+

+

KINEMATICAL EQUATIONS IN RELATIVE FORM (CONSTANT ACCELERATION

When two particles A and B move simultaneously with initial velocitiesu A

and Bu

, at any time 't'

= AAB VV -

BV ;

= BAAB SSS ;

= BAAB aaa

030

045a

x

y

u

030

u

V

ta

045

2

2

1ta

030

tu

S

045


10/60

ABU = BA UU

ABV = tau ABAB

+

2

21 tatuS ABABAB

+=

where

ABX means parameter X of A with respect to B.

Similarly ifr is the position coordinate at time 't' and

0r is the initial

position coordinate at time 't' = 0,

2

10 ++=

turr AAA

2taA

2

10 ++=

turr BBB 2

taB

= ABAB rr 0 +

2

2

1tatu ABAB

+ gives the position coordinate of A with

respect to B at any time.

ABr gives the distance between A and B at any

time 't'.

Illustration 9 :A loose bolt falls from the roof of a lift of height 'h' moving vertically upward withacceleration 'a'. Find the time taken by the bolt to reach the floor of the lift and thevelocity of impact.

Solution :

jhS b =

as the bolt travels a distance 'h' down wards before hitting the

floor

= aaa bb = (-g j ) - (a j ) = - (g + a) j

= uuu bb = ju - ju = 0 as they have the same initial velocity upwards

2

2

1tatuS bbb

+=

- h j = 0 -2

1(a + g) t2j t =

ga

h

+

2

Velocity of impact is nothing but the relative velocity of the bolt with respectto the lift

Vimpact = tauV bbb

+=


11/60

= - (a+g) j ga

h

+

2= ( )gah + 2 j

Illustration 10 :Two particles A and B move on a horizontal surfacewith constant velocities as shown in the figure. If theinitial distance of separation between them is 10 mat t=0, find the distance between them at t = 2 sec

Solution :

Distance between them = ABr

Taking the origin at the initial position of A

20

2

1taturr ABABABAB

++=

ir AB 100 =

( ) ( )jcosicosjcosicosUUu BAAB 0000 301060104521045210 +==

= 5 i - 18.7 j

0==

BAAB aaa

( ) ( )j.iir AB 718510 +=

t

At t = 2 sec , j.r AB 437=

and ABr

= 37.4 units

DISPLACEMENT - TIME GRAPHS :

The displacement is plotted along 'y' axis and the time along 'x' axis. The

slope of the curve

dt

dSgives the instantaneous velocity at that point. The average

slope between two pointst

S

gives the average velocity between these points.

Rate of change of slope gives the acceleration. If the slope is positive anddecreases with time, the particle is under retardation. If the slope is positive andincreases with time, the particle is under acceleration, constant slope implies zeroacceleration.

Illustration 11 :The displacement - time graph of a particle moving along a

straight line is given below. Finda) the time at which the velocity is zero

m10060

s/muB 10=

045

s/muA 210=

A B

0

m2x

2 4 t

circleSemi


12/60

b) the velocity at time t = 1 secc) the average velocity between t = 2 sec and t = 4 sec

Solution :

a) Velocity is zero when the slopoe is zero which happens at t = 2 secb) Since any point (x,t) lies on the circle of radius 2 m and centre (2,0),

(x-0)2 + (t - 2)2 = 22 x = ( ) 224 t

velocity is given by the slopedt

dx= V

V =dt

d ( )

( )( )( )

=

22242

124

2

2t

t

t = +

3

1

Since the slope is +ve between t = 0 and t = 2, v = 3

1

m/s

c) Average velocity =time

ntDisplaceme=

24

2

O= -1 m/s

VELOCITY - TIME GRAPH :

If velocity is plotted on 'y' axis and time is plotted on x axis, the slope of the

curve at any pointdt

dvgiven instantaneous acceleration. The

average slope between two pointst

v

gives average

acceleration. The total area between the curve and the timeaxis gives distance where as algebraic sum of the areas givesdisplacement.

Distance = A1 + A2 + A3Displacement = A1 - A2 + A3

The nature of acceleration can be found from the rate of change of slope

Illustration 12 :The velocity time graph of a particle moving along a straight line has the form of aparabolav = (t2 - 6t + 8) m/s . Finda) the distance travelled between t = 0 second t = 3 sec

b) the velocity of the particle when the acceleration is zero

c) the acceleration of the particle when the velocity is zerod) the velocity of the particle when the acceleration is zero

V1A

2A

3At


13/60

Solution :a) Distance = area OAB + area BCF which can be

obtained by the method of integration.

Since at the points B and D, velocity becomes zero t2 6t + 8 = 0

t = 2 sec and 4secSince F is in between B and D, the time corresponding

to F is2

42 += 3 sec. Similarly A corresponds to t = 0

and E corresponds to t = 6 sec

Area OAB = A1 = ( )

+=+=

2

0

2

0

232

2

0

82

6

386 t

ttdtttdtV =

3

20m

Area BCF = A2 = - mttt

dtV3

28

2

6

3

233

2

=

+=

Distance = m3

22

3

2

3

20=+

b) displacement between t = 3 sec and t = 6 sec = A4 - A3 = A1 - A2 =

3

2

3

20 = 6m

c) a =

dt

dv= 2t - 6

When V = 0 ; t = 2sec and 4 sec

a = 2(2) - 6 and 2(4) - 6= - 2 m/sec2 and 2 m/sec2

d) When a = 0, 2t - 6 = 0 and t = 3 secV = 32 - 6(3) + 8 = -1 m/sec

PROJECTILE MOTION :

At the top most point Vy = 0 and

Vx = u cos

From Vy = uy + ay t, 0 = usin - gt t =

g

sinu

Time of flight = 2t =g

sinu 2

From 22 yy uV = 2ay Sy, 0 - (u sin)2 = 2 (-g) H and H =g

sinu

2

22

Range = (Time of flight) (horizontal velocity) =

( )g

sinucosu

g

sinu =

22 2

R

H

T

u

V

EA

1A

O

B F D 4A

3A2A t

C


14/60

Range is maximum when = 450 and Rmax =g

u2

H

R=

g

sinu

g

cossinu

2

2

22

2

= 4 cot

The velocity of the particle at any time 't' is given by jViVV yx +=

V = (ucos) i + (usin - gt) j

The magnitude of the velocity = ( ) ( )22 gtsinucosu +

If is the angle made by the velocity at any time 't' with the horizontal,

Tan =

cosu

gtsinu

Taking the origin at the point of projection, the 'x' and 'y' coordinates at anytime 't' are given by

x = u cos t and y = usint2

1 gt2

Eliminating 't' from x and y

y = u sin

cosux 21 g

2

cosux

= x tan -22

2

2 cosu

gxwhich is the equation of a parabola.

It may be noted here that the velocity of the projectile will be alwaystangential to its path. The equations of projectile motion derived above arevalid only for constant acceleration due to gravity 'g'.

Illustration 13 :

A particle is projected from the horizontal at an inclination of 600 with an initialvelocity 20 m/sec.

Assuming g = 10 m/sec2 find a) the time at which the energy becomes threefourths kinetic and one fourth potential b) the angle made by the velocity at thattime with the horizontal c) the x and y coordinates of the particle taking theorigin at the point of projection.


15/60

Solution :

a) Let V be the velocity when the given condition is fulfilled2

1mV2 =

4

3

( 2

1

mu2)

V =2

3u= 10 3 m/sec

( ) jgtsinuicosuV +=

= 20 cos 600 i + tsin 106020 0 j = 10 i + jt10310

V = 10 3 102 + ( )210310 t = ( )2310

Solving t = 23 sec

t = 23 while rising up and t = 23 + while coming down

b) Tan =

cosu

gtsinu

=10

2310310 = + 2

c) x = ucos t

= 10 23 = 10 morm 231023 +

and y = u sin t -2

1gt2

= 10 3 ( )23 - 5 ( ) m523 2 =

PROJECTILE MOTION ON AN INCLINED PLANE :

Let be the inclination of the plane and the

particle is projected at an angle with the inclinedplane. It is convenient to take the reference framewith x' along the plane and y' perpendicular to the

plane. gcos will be the component of theacceleration along the downward perpendicular to

the plane and g sin will be the component of the acceleration along the downwarddirection of the inclined plane.

Along the plane, the kinematical equations take the form

'y y

u 'x

cosgsing

x


16/60

tauV 'x'x'x +=

tsingcosuV 'x =

2

2

1tat'uS 'xx'x +=

'xS = ucost -2

1gsin t2

2'xV - 'x'x'x Sau 2

2= 2'xV - (ucos)2 = 2 (-g sin) 'xS

Similarly perpendicular to the plane, the kinematical equations take the form

tauV 'y'y'y += tcosgsinuV 'y =

2

2

1tatuS 'y'y'y +=

2

2

1tcosgtsinuS 'y =

'y'y'y'y SauV 222 = ( ) ( ) 'y'y ScosgsinuV = 2

22

Here it may be noted that,When the particle strikes the inclined plane 'yS = 0

When the particle strikes the inclined plane perpendicular to it, 0='yS and

0='xV

When particle strikes the inclined plane horizontally 0='yS and yV = 0

Illustration 14 :

From the foot of an inclined plane of inclination , a

projectile is shot at an angle with the inclined plane.

Find the relation between and if the projectilestrikes the inclined plane

a) perpendicular to the planeb) horizontally

Solution :

a) Since the particle strkes the plane perpendicularly 0='yS and 0='xV

u sin t -2

1g cos t2 = 0 and u cos - g sin t = 0

t =

cosg

sinu2

and t =

sing

cosu

=

sing

cosu

cosg

sinu2 2 Tan = cot

b) Since the particle strikes the plane horizontally 'yS = 0 and 0=yV

u sin t -2

1g cos t2 = 0 and u sin ( + ) - gt = 0

t =

cosg

sinu2=

( )

g

sinu +

( )

g

sinu

cosg

sinu +=

2

cos

sin2= sin

( + )

'y

u 'x


17/60

CIRCULAR MOTION :

When a particle moves in a circle of radius R with constant speed V, its calleduniform circular motion.

When the particle covers , the direction of velocity also changes by

without change in magnitude. Change in velocity V will be towards the centre of

curvature of the circular path which causes centripetal acceleration. is called theangular position (or) angular displacement.

Centripetal acceleration, ra = t

V

The rate of change of angular position is known as angular velocity ()

Time period of circular motion T =V

R2

In the same time the particle covers an angle 2 from which angular velocity can befound as

=T

2=

R

V

R

V=

2

2

t =

= V

Rand V = + cosVVV 222 2 = 2V sin 2

When is small sin ~ V = V

centripetal acceleration =t

V

=

( )

V

R

V=

R

V2

When speed of the particle continuously changes with time, the

tangential acceleration is given by ta =dt

dV

The rate of change of angular velocity is called the angular acceleration ()

since ra and ta are perpendicular to each other, the resultant

acceleration is given by a = 22 tr aa +

Angle made by the resultant with radius vector Tan =r

t

a

a

V

V

V

V

V


18/60

Illustration 15 :The speed of a particle in circular motion of radius R is given by V = Rt2. Find thetime at which the radial and the tangential accelerations are equal and the distancetraveled by the particle upto that moment.

Solution :

tr aa =

dt

dV

R

V=

2

= 2Rt t = 31

2

Distance travelled = 3

1

2

0

dtV =

3

1

23

3

Rt=

3

2R

RADIUS OF CURVATURE

When a particle is moving in a planeR

Va r

2

= where V is the instantaneous

velocity and R is the radius of curvature at that point.

Radius of curvature =ra

V2

If the path of the particle is given by y = f(x), radius of curvature can also be

found from the formula R =

2

2

2

32

1

dx

yd

dx

dy

+

Illustration 16 :

A particle is projected with initial velocity 'u' at angle with the horizontal. Findthe radius of curvature at a) point of projection b) the top most point

Solution :

a) at the point of projection P, V = u and ra = g cos

R =ra

V2=

cosg

u 2

b) at the topmost point T, V = ucos and ga r =

R =ra

V2=

g

cosu 22

u

p

T


19/60

SHORTEST DISTANCE OF APPROACH :

When two particles A and B are moving simultaneously, their positioncoordinates at any time 't' are given by (when the accelerations are uniform)

20

21

taturr AAAA

++= and 2021

taturr BBBB

++=

The distance between them at any time 't', S = ABr

Where 202

1taturr ABABABAB

++=

The distance between them becomes minimum whendt

dS=0 from which the

time at which it becomes minimum can be found. Substituting the value of

time so obtained in ABr , S min can be found.

Illustration 17 :Two ships A and B move with constantvelocities as shown in the figure. Find theclosest distance of approach between them

Solution :

jrA

100 =

ir B 200 =

jcosicosV A 302060200 =

jcosicosVB 4521045210 +=

ji 31010 = = 10 i + 10 j

0=

Aa 0=

Ba

tVrr AAA

+= 0 tVrr BBB

+= 0

= 10 t i + jt31010 = (20 + 10 t) i + 10 t j

BAAB rrr

= = - 20 i + jtt 1031010

S = ABr

= ( ) ( )22 103101020 tt +

When the distance between A and B is minimumdt

dS= 0

( ) ( )

+22 1031010202

1

tt

1031010310102 tt = 0

A

km10

o30

North

kmphVA 20=

kmphVB 210=

East

045

O km20 B


20/60

10 - 10 3 t - 10 t = 0 t =

+ 31

1hr

Substituting this value of time in the expression for S,S min = 20 km

CYCLIC MOVEMENT OF PARTICLES :

When three or more particles located at the vertices of a polygon of side lmove with constant speed V such that particle 1 moves always towards particle 2and particle 2 moves always towards 3 particle etc., they meet at the centre of thepolygon following identical curved paths.

Time of meeting =approachVelocityof

seperationInitial

Velocity of approach is the component of the relative velocity along the linejoining the particles.

Illustration 18 :

Six particles located at the six vertices of a hexagon of side l move with constantspeeds V such that each particle always targets the particle in front if it. Find thetime of meeting and the distance travelled by each particle before they meet

Solution :

t = approachofVelocity

seperationInitial

=060cosVV

=

V

2

Since they move with constant speed V, the distance travelled by

each particle in time t =V

2is d = Vt = V

V

2= 2 l

RIVER PROBLEMS :

If rV

is the velocity of the river and bV

is the

velocity of the boat with respect to still water,the resultant velocity of the boat RV

= rb VV

+

Only the perpendicular component of theresultant velocity helps in crossing the river.

Time of crossing, t =cosV

w

b

where 'w' is the

width of the river.

The boat crosses the river in the least time when = 0

The parallel component of the resultant velocity determines the drift.

Drift is the displacement of the boat parallel to the river by the time the boatcrosses the river

V

V

V

V

V

060

B

rV

A

C

bV


21/60

Drift , x = ( ) sinVV br

cosV

w

b

Zero drift is possible only when Vr = Vb sin. When Vr > Vb zero drift is not

possible.

Illustration 19 :A river of width 100 m is flowing towards East with a velocity of 5 m/s. A boatwhich can move with a speed of 20 m/s with respect to still water starts from a pointon the South bank to reach a directly opposite point on the North bank. If a wind is

blowing towards North East with a velocity of 5 2 m/s, find the time of crossingand the angle at which the boat must be rowed.

Solution :

jcosisinV b +=

2020

rV

= 5 i

V w = 5 2 cos 45 i + 5 2 cos 45 j = 5 i + 5 j

RV

= Resultant velocity of the boat = bV

+ rV

+ wV

= ( - 20 sin + 5 + 5) i + (20 cos + 5) j

For reaching directly opposite point, the component of the resultant velocity

parallel to the river must be zero

- 20 sin + 10 = 0 sin =2

1and = 300

Since time of crossing depends only on the perpendicular component of theresultant velocity.

t =520 +cos

w=

53020

1000 +cos

= 4.48 sec

rVwV

45

bV


22/60

WORKED OUT OBJECTIVE PROBLEMS

EXAMPLE : 01A point moves along 'x' axis. Its position at time 't' is given by x2 = t 2 + 1. Its

acceleration at time 't' is

A)3

1

xB)

2

11

xx C)

2x

tD)

3

2

x

t

Solution :

x = 12 +t ;dt

dx=

12

12 +t

(2t) =12 +t

t

a =2

2

xt

xd=

22

2

2

1

2121

+

+

+

t

)t(

t

tt

=

( )32 11

+t

=3

1

x

EXAMPLE : 02

A body thrown vertically up from the ground passes the height 10.2m twice at aninterval of 10 sec. Its initial velocity was (g = 10 m/s2)A) 52 m/s B) 26 m/s C) 35 m/s D) 60m/sSolution :

Displacement is same in both casess = ut + 1/2 at2

10.2 = ut -2

1(10) t2 t =

10

2042 uu

t1 =10

2042 uuand t2 =

10

2042 + uu t = t2 - t1 = 10 sec

2042 u = 50 u2 = 2500 + 204 u = 52 m/s

EXAMPLE : 03A car starts from rest moving along a line, first with acceleration a= 2 m/s2, thenuniformly and finally decelerating at the same rate and comes to rest. The totaltime of motion is 10 sec. The average speed during this time is 3.2 m/s. How longdoes the car move uniformlyA) 4 sec B) 6 sec C) 5 sec D) 3sec

Solution :


23/60

Let the car accelerate for time 't' and move uniformly with v = at for time t1Since the magnitudes of acceleration and deceleration are same, the time ofdeceleration is also 't'.

t + t 1 + t = 10 sec

Average speed =time

Distance=

( )

10

21

21 2

12

++

attatat

= 3.2

2t2 + 2tt1 = 32 2 322

102

2

101

12

1 =

+

t

tt

Solving t1 = 6 sec This problem can be solved using velocity time graphalso.

EXAMPLE : 04A particle has an initial velocity of ( )ji 43 + m/s and a constant acceleration( )ji 34 m/s2. Its speed after 1 sec will be equal toA) zero B) 10 m/s C) 5 2 m/s D) 25m/s

Solution :

tauV

+= = ( ) ( )jiji 3443 ++ (1) = 7 i - j

Speed = magnitude of

V=

22

17+

= 5 2 m/s

EXAMPLE : 05

An aeroplane flies along a straight line from A to B with air speed V and backagain with the same air speed. If the distance between A and B is l and a steadywind blows perpendicular to AB with speed u, the total time taken for the roundtrip is

A)V

2B)

22

2

uV +

C)

22

2

uV

VD)

222

uV

Solution :The resultant velocity of the plane must be along AB during forward

journey.

t1 =22 uVVR

=

During return journey, the resultant velocity of the plane must be along

BA

A

RV

B

V

u

RV

V

A

Bu


24/60

t2 =22 uVVR

=

Total time t = t1 + t2 =22

2

uV

EXAMPLE : 06

A particle is thrown with a speed 'u' at an angle with the horizontal. When the

particle makes an angle with the horizontal its speed changes to V. Then

A) V = u cos B) V = ucos cos C) V = u cos sec D) V = usec

cos

Solution :

Since the horizontal component of the velocity of a projectile always remainsconstant

u cos=V cos V=ucos sec

EXAMPLE : 07

Two shells are fired from a cannon with same speed at angle and respectivelywith the horizontal. The time interval between the shots is T. They collide in midair after time 't' from the first shot. Which of the following conditions must besatisfied.

A) > B) t cos = (t -T) cos

C) (t-T) cos=cos D) (usin)t - 21gt2 =(usin) (t-T)- 21

g(t-T)2

Solution :When they collide, their 'x' and 'y' components must be sameucos t = u cos (t-T) cos t = cos (t-T)

(usin) t -2

1gt2 = (usin) (t-T) -

2

1g (t-T)2

Since cos = cos

t

T1 and T < t

cos < cos and >

EXAMPLE : 08

A particle is projected from a point 'p' with velocity 5 2 m/s perpendicular to thesurface hollow right angle cone whose axis is vertical. It collides at point Qnormally on the inner surface. The time of flight of the particle is

A) 1 sec B) 2 sec C) 2 2 sec D) 2sec

Solution :

It can be seen from the diagram thatV becomes perpendicular to

u .


25/60

u = ucos450 i + u sin45 j

V =

u +

a t = (ucos 45 i + usin45 j ) - (gt) j

When

V becomes perpendicular to

u ,

V .

u = 0

u2 cos2 45 + u2 sin2 45 - (usin45) gt = 0 t =45sing

u= 1 sec

EXAMPLE : 09A man walking Eastward at 5 m/s observes that the wind is blowing from theNorth. On doubling his speed Eastward he observes that the wind is blowingfrom North East. The velocity of the wind isA) (5i+5j) m/s B) (5i - 5j) m/s C) (-5i +5j) m/s D) (-5i -5j) m/s

Solution :

let jViVV w 21 +=

In the first case mwwm VVV

= = ( )jViV 21 + - ( )i5Since no component along East is observed V1 - 5 = 0 V1 = 5 m/sIn the second case

mwwm VVV

= = (V1 i + V2j ) - (10 i ) = ( ) jViV 21 10 +

Since the wind is observed from North East the components along North andEast must be same V1 - 10 = V2 V2 = - 5 m/s

V w = (5i - 5j) m/s

EXAMPLE :10From a lift moving upward with uniform acceleration 'a', a man throws a ballvertically upwards with a velocity V relative to the lift. The time after which itcomes back to the man is

A)ag

V

2B)

ag

V

+C)

ag

V

+

2D)

222

ag

Vg

Solution :

Since the velocity of the ball is given relative to the lift blV

= V j

When the ball comes back to the man, its displacement relative to the lift is

zero blS

= 0

bla

= lb aa

= (-g) j - a j = - (g + a) j

Applying S = ut + 1/2 at2 in relative form


26/60

blbl VS

= t +2

1 bla

t2

0 = ( )jVt +

2

1 ( )( )jag + t2 t =

ag

V

+

2


27/60

ASSIGNMENT

SINGLE ANSWER TYPE QUESTIONSLEVEL I1. The greatest acceleration or deceleration that a train may have is a. The

minimum time in which the train can go from one station to the next at adistance S is

A)a

SB)

a

S2C) 2

a

sD)

a2

s

2. A car accelerates from rest at a constant rate for sometime and attains a

velocity of 20 m/s. Afterwards it decelerates with a constant rate /2 andcomes to a halt. If the total time taken is 10s, the distance travelled by the carisA) 200m B) 100m C) 10m D) 20m

3. A particle starts from the position of rest under a constant acceleration. It

travels a distance x in the first 10 seconds and distance y in the next 20seconds. ThenA) y = x B) y = 2x C) y = 8x D) y = 4x

4. A body is projected vertically upwards. If t1 and t2 be the times at which it isat height h above the point of projection while ascending and descendingrespectively, then h is

A) 21ttg2

1B) g t1 t2 C) 2 g t1 t2 D) 4 g t1 t2

5. From a 20m high tower one ball is thrown upward with speed of 10m/s andanother is thrown vertically downward at the same speed simultaneously.

The time difference of their reaching the ground will be (take g = 10m/s2)A) 12s B) 6s C) 2s D) 1s

6. A particle X moving with a constant velocity u crosses a point O. At the sameinstant another particle F starts from rest from O with a constant accelerationa. The maximum separation between them before they meet is

A)a2

u2

B)a

u2

C)a

u22

D)a4

u2

7. A bird flies in straight line for 4s with a velocity v = (2t-4) m/s. What is thedistance covered by the bird in returning to the place from where it started its

journey ?A) 0 B) 8m C) 4m D) 2m

8. A ball is thrown vertically upwards. Which of the following plots representsthe speed-time graph of the ball during its flight if the air resistance is notignored

A) B) C) D)


28/60

9. If a ball is thrown vertically upwards with speed 'u', the distance coveredduring the last t second of its ascent is

A) (u+gt)s B) ut C)2

1gt2 D) ut -

2

1gt2

10. A particle has an initial velocity of 9m/s due east and a constant accelerationof 2m/s2 due west. The distance covered by the particle in the fifth second ofits motion isA) 0 B) 0.5m C) 2m D) none ofthese

11. Two particles are projected simultaneously in the same vertical plane fromthe same point, with different speeds u1 and u2, making angles 1 and 2

respectively with the horizontal , such that u1 cos1 = u2cos2. The pathfollowed by one, as seen by the other (as long as both are in flight), is :A) a horizontal straight line B) a vertical straight line C) a parabola

D) a straight line making an angle |1 - 2| with the horizontal

12. A particle starts from the origin of coordinates at time t = 0 and moves in thexy plane with a constant acceleration in the y-direction. Its equation of

motion is y =x2. Its velocity component in the x-direction is

A) variable B)

2C)

2D)

2

13. A train starts from station A with uniform acceleration for some distance

and then goes with uniform retardation for some more distance to come to

rest at station B. The distance between station A and B is 4 km and the traintakes 4 minute to complete this journey. If and are in km (min)-2 then

A) 211=

+

B) 4

11=

+

C)

2

111=

+

D)

4

111=

+

14. The driver of a train moving with a speed v1 sights another train at a distanced, ahead of him moving in the same direction with a slower speed v2. Heapplies the breaks and gives a constant de-acceleration 'a' to his train. For nocollision, d is

A) =( )

a2

vv2

21 B) >( )

a2

vv2

21 C) t2

B) R' < R, H' < H, '1t > t1 and'2t < t2

C) R' < R, H' > H, '1t > t1 and'2t < t2

D) R' < R, H' < H, '1t < t1 and'2t > t2

35. Speed of a particle moving in a circle varies with time as, v = 2t. Then :A) angle between velocity vector and acceleration vector is increasing withtime.

B)a

is constant while ar is increasing with time.C) both A and B are correct D) both A and B are wrong.36. Initial velocity and acceleration of two particles are as shown in fig. Assuming

the shown direction as the positive, vBA versus time graph is as :

A) B) C) D)

37. A graph is plotted between velocity (v) and displacement (s) of particlemoving in a straight line. Here v is plotted along y-axis and 's' along x-axis.Choose the correct option.A) slope of this graph at any point always gives us the ratio of velocity anddisplacement at that point.7

B) slope represents a/v under all the conditions. (a = acceleration)C) both A and B are correct D) both A and B are wrong.


32/60

38. In a projectile motion if a person wants to increase the maximum height to 2times but simultaneously want to decrease the range same number of time.He can achieve it by increasing tan of angle of projection by .......times.

A) 2 B) 4 C) 3 D) 2 39. The velocity of a particle moving in a straight line varies with time

in such a manner that v versus t graph is represented by one halfof an ellipse. The maximum velocity is m and total time of motionis t0i) Average velocity of particle is /4m ii) Such motion can not be realized inpractical termsA) Only (i) is correct B) Only (ii) is correctC) Both (i) and (ii) are correct D) Both (i) and (ii) are wrong

40. Starting from rest, a particle rotates in a circle of radius R = 2 m with an

angular acceleration = /4 rad/ s2. The magnitude of average velocity of theparticle over the time it rotates quarter circle isA) 1.5 m/s B) 2 m/s C) 1 m/s D) 1.25m/s

41. In a car race car A takes t0 time less to finish than car B and passes thefinishing point with a velocity v0 more than car B. The cars start from rest and

travel with constant accelerations a1 and a2. Then the ratio0/t0 is equal to

A)2

21

a

aB)

2

aa 21 + C) 21 aa D)1

22

a

a

42. A rod of length 1 leans by its upper end against a smooth vertical

wall, while its other end leans against the floor. The end that leansagainst the wall moves uniformly downward. ThenA) The other end also moves uniformlyB) The speed of other end goes on decreasingC) The speed of other end goes on increasingD) The speed of other end first decreases and then increases

43. A particle is moving along a circular path of radius 5 m and with uniformspeed 5 m/s. What will be the average acceleration when the particlecompletes half revolution?

A) zero B) m/s2 C) 10 m/s2 D) 10/m/s2

44. The velocity displacement graph of a particle moving along a straightline is shownThe most suitable acceleration-displacement graph will be

A) B)


33/60

C) D)

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

C B C A C A B C C B B D A B B A A A B B21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40

B C A C C B A C A A C D B D C A B B C C41 42 43 44

C A

LEVEL II1. A particle starts from rest at time t = 0 and moves

on a straight line with an acceleration which varieswith time as shown in fig. The speed of the particlewill be maximum after how many seconds

A) 4s B) 6sC) 8s D) 10s

2. Due to air a falling body faces a resistive force proportional to square ofvelocity v, consequently its effective downward acceleration is reduced and isgiven by a = g - kv2 where k = 0.002m-1. The terminal velocity of the falling

body isA) 49m/s B) 70m/s C) 9.8m/s D) 98m/s

3. A balloon is rising with a constant acceleration of 2m/s2. At a certain instantwhen the balloon was moving with a velocity of 4m/s, a stone was droppedfrom it in a region where g = 10m/s2. The velocity and acceleration of stone asit comes out from the balloon are respectively.A) 0, 10m/s2 B) 4m/s, 8m/s2 C) 4m/s, 12m/s2 D) 4m/s,

10m/s2

4. A stone is thrown vertically up from the top of a tower with some initialvelocity and it arrives on the ground after t1 seconds. Now if the same stone isthrown vertically down from the top of the same tower with the same initialvelocity, it arrives on ground after t2 seconds. How much time will the stonetake to reach the ground if it is dropped from the same tower ?

A)2

tt 21 + B)2

tt 21 C) 21 tt + D) 21tt

5. Two particles A and B start from the same point and slide down throughstraight smooth planes inclined at 300 and 600 to the vertical and in the same

vertical plane and on the same side of vertical drawn from the starting point.The acceleration of B with respect to A is


34/60

A) g/2 in vertical direction B) g 2/3 at 450 to vertical

C) g/ 3 at 600 to vertical D) g in vertical direction

6. A particle starts from rest at the origin and moves along X-axis withacceleration a = 12-2t. The time after which the particle arrives at the origin isA) 6s B) 18s C) 12s D) 4s

7. Figure represents position (x) versus time (t)graph for the motion of a particle. If b and c are

both positive constants, which of the followingexpressions best describes the acceleration (A)of the particle ?A) a = +bB) a = -cC) a = b + ctD) a = b-ct

8. Two particles instantaneously at A and B are 5m, apartand they are moving with uniform velocities, theformer towards B at 4m/s and the latter perpendicularto AB at 3m/s. They are nearest at the instantA) 2/5sB) 3/5sC) 1sD) 4/5s

9. Three particles start from the origin at the same time, one with a velocity 'u1',alone X-axis, the second along the Y-axis with a velocity u2 and the thirdalogn x = y line. The velocity of third particle so that the three may always lieon the same plane is

A)2

uu 21 + B) 21uu C)21

21

uu

uu

+D)

21

21

uu

uu2

+

10. A ball is shot vertically upwards from thesurface of a planet in a distant solar system.A plot of the y versus t for the ball is shownin fig. The magnitude of the free fall in m/s2

on the planet isA) 4B) 8C) 12D) 16

11. The horizontal range of a projectile is R and the maximum height attained byit is H. A strong wind now begins to blow in the direction of the motion of theprojectile, giving it a constant horizontal acceleration = g/2. Under the sameconditions of projection, the horizontal range of the projectile will now be

A) R +2H B) R + H C) R +

2H3 D) R + 2H


35/60

12. A particle moves int he xy plane with a constant acceleration g in the negativey-direction. Its equation of motion is y = ax - bx2, where a and b are constants.Which of the following are correct ?A) The x-component of its velocity is constant

B) at the origin, the y-component of its velocity is ab2

g

C) At the origin, its velocity makes an angle tan-1(a) with the x-axis.D) the particle moves exactly like a projectile.

13. Two bodies are projected simultaneously from the same point, in the samevertical plane, one towards east and other towards west with velocities 8 ms -1and 2 ms-1 respectively. The time at which their velocities are perpendicular toeach other isA) 2/5 s B) 5/2s C) 1/5 s D) 5 s

14. Two stones are projected so as to reach the same distance from the point of

projection on a horizontal surface. The maximum height reached by oneexceeds the other by an amount equal to half the sum of the heights attained

by them. Then the angles of projection for the stones areA) 450, 1350 B) 00, 900 C) 300, 600 D) 200, 700

15. Velocity and acceleration of a particle at some instant of time are ( )j4i3v += m/s and ( )j8i6a += m/s2 respectively. At the same instant particle is atorigin. Maximum x-co-ordinate of particle will beA) 1.5m B) 0.75m C) 2.25m D) 4.0m

16. a-t graph for a particle moving in a straight line is as

shown in figure. Change in velocity of the particlefrom t=0 to t=6s is :A) 10m/s B) 4m/sC) 12m/s D) 8m/s

17. Speed time graph of two cars A and B approachingtowards each other is shown in figure. Initialdistance between them is 60m. The two cars willcross each other after time.

A) 2sec B) 3secC) 1.5sec D) 2 sec

18. The position of a particle along x-axis at time t is given by x = 2 + t - 3t2. Thedisplacement and the distance travelled in the interval t = 0 to t = 1 arerespectivelyA) 2, 2 B) -2, 2.5 C) 0, 2 D) -2, 2.16

19. The acceleration time graph of a particle movingalong a straight line is as shown in figure. At what

time the particle acquires its initial velocity ?A) 12s


36/60

B) 5sC) 8sD) 16s

20. A graph between the square of the velocityof a particle and the distance s moved bythe particle is shown in the figure. Theacceleration of the particle in kilometre perhour square is :A) 2250B) 225C) -2250D) -225

21. A particle starts from rest and traverses a distance l with uniform acceleration,then moves uniformly over a further distance 2l and finally comes to rest aftermoving a further distance 3l under uniform retardation. Assuming entiremotion to be rectilinear motion the ratio of average speed over the journey tothe maximum speed on its way is :A) 1/5 B) 2/5 C) 3/5 D) 4/5

22. Two stones are thrown up simultaneously with initial speeds of u1 andu2 (u2 >u1). They hit the ground after 6s and 10s respectively. Which graph in figure

correctly represents the time variation ofx = (x2 - x1), the relative position ofthe second stone with respect to the first upto t = 10s ? Assume that the stones

do not rebound after hitting the ground.

A) B) C) D)

23. Figure shows the position-time (x-t) graph of the motionof two boys A and B returning from their school O totheir homes P and Q respectively. Which of thefollowing statements is true ?A) A walks faster than BB) Both A and B reach home at the same timeC) B starts for home earlier than A

D) A overtakes B on his way to home


37/60

24. A ball is projected with a velocity 20 3 m/s at angle 600 to the horizontal. The

time interval after which the velocity vector will make an angle 300 to thehorizontal is (take g = 10m/s2)A) 4sec B) 2 sec C) 1 sec D) 3 sec

25. The equation of motion of a projectile is : y = 12x -4

3x2

Given that g = 10ms-2, what is the range of the projectile ?A) 12m B) 16m C) 20m D) 24m

26. A projectile is thrown with an initial velocity of ( ) 1msjbia + . If the range ofthe projectile is twice the maximum height reached by it, then :A) a = 2b B) b=a C) b = 2a D) b = 4a

27. A particle moves along a parabolic path y = 9x2 in such a way that the x

component of velocity remains constant and has a value 1ms3

1 . The

acceleration of the particle is

A) 2msj3

1 B) 2msj3 C) 2msj3

2 D) 2msj2

28. Two projectiles are projected with the same velocity. If one is projected at anangle of 300 and the other at 600 to the horizontal. The ratio of maximum

heights reached, is :A) 1 : 3 B) 2 : 1 C) 3 : 1 D) 1 : 4

29. A particle is projected from the ground with velocity u at angle withhorizontal. The horizontal range, maximum height and time of flight are R, Hand T respectively. They are given by,

R =g2

sinuH,

g

2sinu222

=

and T =g

sinu2

Now keeping u as fixed, is varied from 300 to 600. Then,A) R will first increase then decrease, H will increase and T will decreaseB) R will first increase then decrease, while H and T both will increase

C) R will decrease while H and T will increaseD) R will increase while H and T will decrease

30. Velocity and acceleration of a particle at some instant of time are

( )k2ji2v += m/s and ( )kj6ia += m/s2 . Then, the speed of the particle is....... at a rate of ....... m/s2A) increasing, 2 B) decreasing ,2 C) increasing, 4 D)decreasing, 4

31. x and y coordinates of a particle moving in xy plane at some instant are :

x = 2t2 and y = 2t2

3


38/60

The average velocity of particle in a time interval from t = 1 second to t = 2second is :

A) ( ) s/mj5i8 + B) ( ) s/mj9i12 + C) ( ) s/mj5.4i6 + D)

( )s/mj6i10 +

32. A particle is projected upwards with some velocity. At what height fromground should another particle be just dropped at the same time so that bothreach the ground simultaneously. Assume that first particle reaches to amaximum height H.A) 6H B) 8H C) 4H D) 10H

33. Particle A moves with 4m/s along positive y-axis and particle B in a circle x2+ y2 = 4(anticlockwise) with constant angular velocity = 2 rad/s. At time t =0 particle is at (2m, 0). Then :

A) magnitude of relative velocity between them at time t is 8 sin 2tB) magnitude of relative velocity between them is maximum at t = /4second.C) both A and B are correctD) both A and B are wrong

34. An armored car 2 m long and 3 m wide is moving at 10 ms-1 when a bullet hitsit in a direction making an angle tan-1(3/4) which the length of the car as seen

by a stationary observer. The bullet enters one edge of the car at the cornerand passes out at the diagonally opposite corner. Neglecting any interaction

between the car and the bullet, the time for the bullet to cross the car isA) 0.20 s B) 0.15 s C) 0.10 s D) 0.50 s

35. The V - t graph for the rectilinear motion of a particle is represented by aparabola as shown in fig. Find the distance traveled by the particle in timeT/2.

A) 3

TV2max

B) 2

TV2max

C) 3

TVmax

D) 2

TV 2max

36. A glass wind screen whose inclination with the vertical can be changed ismounted on a car. The car moves horizontally with a speed of 2 m/s. At whatangle with the vertical should the wind screen be placed so that the raindrops falling vertically downwards with velocity 6 m/s strike the wind screenperpendicularly.A) tan-1 (1/3) B) tan-1 (3) C) cos-1(3) D) sin-1(1/3)

37. Two stones are thrown up simultaneously from the edge of a cliff with initialspeeds v and 2 v. The relative position of the second stone with respect to firstvaries with time till both the stones strike the ground asA) Linearly B) First linearly then parabolically

C) Parabolically D) First parabolically then linearly


39/60

38. There are two values of time for which a projectile is at the same height. Thesum of these two times is equal toA) 3T/2 B) 4T/3 C) 3T/4 D) T

39. A particle is projected from a horizontal plane with 8 2 m/s at an angle. At

highest point its velocity is found to be 8 m/s. Its range will beA) 6.4 m B) 3.2 m C) 5 m D) 12.8 m

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

A B D D A B D D D B D A A C B B B D C C21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39

C A B B B C D A B B C C D A C D


40/60

LEVEL - III1. Two particles A and B are connected by a rigid rod AB.

The rod slides on perpendicular rails as shown in fig. Thevelocity of A to the left is 10m/s. What is velocity of B

when = 600?A) 10m/sB) 5.8m/sC) 17.3m/s D) 9.8m/s

2. In the figure , the pulley P moves to the rightwith a constant speed u. The downwardspeed of A is vA, and the speed of B to theright is vB.A) vB = vA

B) vB = u + vAC) vB + u = vAD) The two blocks have accelerations of the same magnitude

3. A marble starts falling from rest on a smooth inclined plane forming an angle

with horizontal. After covering distance 'h' the ball rebound off the plane.The distance from the impact point where the ball rebounds for second time is

A) 8h cos B) 8h sin C) 2h tan D) 4h sin4. From the top of a tower of height 40m, a ball is projected upwards with a

speed of 20m/s at an angle of elevation of 300. The ratio of the total time takenby the ball to hit the ground to its time of flight (time taken to come back to

the same elevation) is (take g = 10m/s2)A) 2 : 1 B)3 : 1 C) 3 : 2 D) 1.5 : 1

5. If time taken by the projectile to reach Q is T, then PQ is equal to :

A) Tv sin

B) Tv cos

C) Tv sec

D) Tv tan

6. A particle is thrown with a speed u at an angle with the horizontal. Whenthe particle makes an angle with the horizontal, its speed changes to v :

A)v = u cos B) v = u cos cos C) v = u cos sec D) v = u sec

cos

7. A stone is projected from a point on the ground so as to hit a bird on the topof a vertical pole of height h and then attain a maximum height 2h above theground. If at the instant of projection the bird flies away horizontally with auniform speed and if the stone hits the bird while descending then the ratio ofthe speed of the bird to the horizontal speed of the stone is :

A)12

2+

B)12

2

C)21

21 + D)

122+


41/60

8. Shots are fired simultaneously from the top and bottom of

a vertical cliff with the elevation = 300, = 600respectively and strike the object simultaneously at thesame point.

If a = 30 3 m is the horizontal distance of the object from

the cliff, then the height of the cliff is :A) 30m B) 45mC) 60m D) 90m

9. A particle if projected up an inclined plane of length 20m and inclination 300(with horizontal). What should be the value of angle (with horizontal) withwhich the projectile be projected so that it strikes the plane exactly at mid-point, horizontally :

A) = tan-1

3

2B) = tan-1 (2) C) = tan-1 ( 3 ) D) = tan-1

2

3

10. A stone is projected at an angle with the horizontal with velocity u. Itexecutes a nearly circular motion near its maximum height for a short time.The radius of circular path is :

A)g2

cosu 22 B)

g

sinu 22 C)

g

cosu 22 D)

g

u 2

11. Two projectiles A and B are fired simultaneously asshown in figure. They collide in air at point P at time t.then :

A) t(u1 cos1 - u2 cos2) = 20

B) t(u1 sin1 - u2 sin2) = 10C) both (A) and (B) are correct

D) both (A) and (B) are wrong

12. Two stones are projected simultaneously with equal speeds from point on aninclined plane along the line of its greatest slope upwards and downwardsrepectively. The maximum distance between their points of striking the planeis double that of when they are projected on a horizontal ground with samespeed. If one strikes the plane after two seconds of the other, the angle ofinclination of plane isA) 300 B) 450 C) 350 D) 150

13. A particle is projected under gravity with velocity ga2 from a point at aheight h above the level plane. The maximum range R on the ground is


42/60

A) h)1a( 2 +

B) ha 2

C) ha

D) 2 )ha(a +

14. The co-ordinates of a particle moving in a plane are given by x = a cos pt andy = b sin pt where a, b (< a) and p are positive constants of appropriatedimensions. ThenA) The path of the particle is an ellipseB) The velocity and acceleration of the particle are normal to each other at t =

/2pC) The acceleration of the particle is always directed towards a fixed point

D) The distance traveled by the particle in time interval t = 0 to t = /2p is a.

1 2 3 4 5 6 7 8 9 10 11 12 13 14

B BD B A D C D C A C B B D

MULTIPLE ANSWER TYPE QUESTIONS

1. A particle of mass m is thrown up vertically with velocity u. As air exerts a

constant force F, the particle returns back at the point of projection with velocityv after attaining maximum height h, then

A))m/Fg(2

uh

2

+= B)

)m/Fg(2h

2

=

vC)

)m/Fg(

)m/Fg(u

+

=v D)

)m/Fg(

)m/Fg(u

+=v

2. A particle of mass m moves on X-axis as follows; it starts from rest at t = 0from the point x = 0, and comes to rest at t = 1 at the point x = 1. No otherinformation is available about its motion at intermediate times (0 < t < 1). Ifdenotes the instantaneous acceleration of the particle, then

A) cannot remain positive for all t in the interval 0 t 1

B) || cannot exceed 2 at any point in its path

C) || must be 4 at some point or points in its path

D) must change sign during the motion, but no other assertion can be madewith the information given

3. Two trains are travelling along a straight track one behind the other. The firsttrain is travelling at 12 m/s. The second train, approaching from the rear istravelling at speed v > 12 m/s when the second train is 200 m behind the first,the driver of second train applies brakes producing a uniform deceleration of

0.20 m/s2. ThenA) If v = 20 m/s, the trains will not collide

V= ag2


43/60

B) If v = 20 /s, the trains will collide after about 20 s

C) If v = 27 m/s, the trains will not collide

D) If v = 27 m/s, the trains will collide after about 15s

4. A particle of mass m and charge q starts from rest from origin along X-axis in

a region where an electric field E = E0 - a x exists. Here E0 and a are constantand x is the distance from the starting point. Then in the region between x = 0to x = 2 E0/a.

A) the speed of particle first increases, then decreases

B) the particle comes to rest at x = 2 E0/a

C) the particle has maximum speed at x = E0/a

D) the particle is subjected to an acceleration which changes sign at x = E0/a

5. Two particles are projected from the same point with the same speed v0 at thedifferent angles 1 and 2 with the horizontal. Their respective times of

flights are T1 and T2. If they have the same horizontal range, and theirmaximum heights are H1 and H2 respectively, then

A) 1 + 2 = 900 B) H1 + H2 = v02/2g C) 12

1 tanT

T= D)

12

2

1 tanH

H=

6. A projectile thrown on a level surface attains a height h after t1 seconds andagain after t2 seconds. If the maximum height attained by the projectile is Hafter t seconds, then

A) t1t2 = g

h2

B) t1 + t2 = g

H8

C) t2 - t1 = g

)hH(8

D) t2 - t = t - t1

7. Which of the following statements are true for a moving body?

A) If its speed changes, its velocity must change and it must have someacceleration

B) If its velocity changes, its speed must change and it must have someacceleration

C) If its velocity changes, its speed may or may not change, and it must havesome acceleration

D) If its speed changes but direction of motion does not change, its velocity

may remain constant8. The figure shows the velocity () of a particle plotted

against time (t)A) The particle changes its direction of motion atsome pointB) The acceleration of the particle remains constantC) The displacement of the particle is zeroD) The initial and final speeds of the particle are the same

9. A particle starts from the origin of coordinates at time t = 0 and moves in the

xy plane with a constant acceleration in the y-direction. Its equation ofmotion is y = x2. Its velocity component in the x-direction is


44/60

A) variable B)

2C)

2D)

2

10. Two particles A and B start simultaneously from the same point and move ina horizontal plane. A has an initial velocity u1 due east and acceleration a1 due

north. B has an initial velocity u2 due north and acceleration a2 due eastA) Their paths must intersect at some pointB) They must collide at some pointC) They will collide only if a1u1 = a2u2D) If u1 > u2 and a1 < a2, the particles will have the same speed at some pointof time

11. Two particles are projected from the same point with the same speed, atdifferent angels 1 and 2 to the horizontal. They have the same horizontalrange. Their times of flight are t1 and t2 respectively

A) 1 + 2 = 900

B) 121

tant

t

= C) 221

tant

t

= D)

2

2

1

1

sin

t

sin

t

=

12. A large rectangular box falls vertically with an acceleration a. Atoy gun fixed at A and aimed towards C fires a particle P.A) P will hit C if a = g B) P will hit the roof BC if a > gC) P will hit the wall CD or the floor AD if a < gD) May be either A, B, or C depending on the speed of projectionof P

13. Two shells are filed from a cannon with speed u each, at angles of and respectively with the horizontal. The time interval between the shots is T.They collide in mid air after time t from the first shot. Which of the followingconditions must be satisfied?

A) > B) t cos = (t - T) cos C) (t - T) cos = t cos

D) (usin ) t -2

1gt2 = (u sin ) (t - T) -

2

1g (t - T)2

14. A particle moving with a speed v changes direction by an angle , withoutchange in speedA) The change in the magnitude of its velocity is zero

B) The change in the magnitude of its velocity is 2v sin /2

C) The magnitude of the change in its velocity is 2 v sin /2

D) The magnitude of the change is its velocity is v (1 - cos )15. A particle of mass 'm' moves on the x-axis as follows:

It starts from rest at t = 0 from the point x = 0 and comes to rest at t = 1 at thepoint x = 1.No other information is available about its motion at intermediate times ( 0 < t< 1)

If denotes the instantaneous acceleration of the particle, thenA) cannot remain positive for all t in the interval 0 t 1


45/60

B) cannot exceed 2 at any point on its path

C) must be 4 at some point or points in its path

D) must change sign during the motion, but no other assertion can be madewith the information given

16. A point moves such that its displacement as a function of time is given by x2 =t2 + 1. Its acceleration at time t is

A)3x

1B)

2x

t C)

3

2

x

t

x

1 D)

2x

t

x

1

17. A body falls from a large height 'h' in 't' second. The time taken to cover thelast metre is

A)gh

1B)

gh2

1C)

gt

1D)

gt2

1

18. A bead is free to slide down a smooth wire tightly

stretched between the points P1 and P2 on a vertical circleof radius R. If the bead starts from rest from P1, thehighest point on the circle and P2 lies anywhere on thecircumference of the circle. Then,A) time taken by bead to go from P1 to P2 is dependent on

position of P2 and equals 2g

Rcos

B) time taken by bead to go from P1 to P2 is independent of position of P2 and

equals 2g

R

C) acceleration of bead along the wire is g cos

D) velocity of bead when it arrives at P2 is 2 gR cos

19. The position of a particle travelling along x axis is given by xt = t3 - 9t2 + 6twhere xt is in cm and t is in second. Then

A) the body comes to rest firstly at (3 - 7 ) s and then at (3 + 7 )s

B) the total displacement of the particle in travelling from the first zero ofvelocity to the second zero of velocity is zeroC) the total displacement of the particle in travelling from the first zero ofvelocity to the second zero of velocity is -74 cm

D) the particle reverses its velocity at (3 - 7 )s and then at (3 + 7 )s and has anegative velocity for (3 - 7 ) < t < (3 + 7 )

20. The velocity of a particle moving along a straight line increases according tothe linear law v = v0 + kx, where k is a constant. ThenA) the acceleration of the particle is k(v0 + kx)

B) the particle takes a time

0

1e

v

vlog

k

1to attain a velocity v1

C) velocity varies linearly with displacement with slope of velocitydisplacement curve equal to kD) data is insufficient to arrive at a conclusion


46/60

21. A particle moves with an initial velocity v0 and retardation v, where v isvelocity at any istant t. Then

A) the particle will cover a total distance

0v

B) the particle continues to move for a long time span

C) the particle attains a velocity 0v2

1at t =

1

D) the particle comes to rest at t =

1

22. An aeroplane flies along a straight line from A to B with a speed v0 and backagain with the same speed v0. A steady wind v is blowing. If AB = l then

A) total time for the trip is22

0

0

vv

v2

l, if wind blows along the line AB

B) total time for the trip is 220 vv

2

l

, if wind blows perpendicular to the line

ABC) total time for the trip decreases because of the presence of windD) total time for the trip increases because of thepresence of wind

23. At the instant a motor bike starts from rest in agiven direction, a car overtakes the motor bike,

both moving in the same direction. The speed timegraphs for motor bike and car are represented by

OAB and CD respectively. ThenA) at t = 18 s the motor bike and car are 180 mapartB) at t = 18 s the motor bike and car are 720 m apartC) the relative distance between motor bike and car reduces to zero at t = 27 sand both are 1080 m far from originD) the relative distance between motor bike and car always remains same

24. A particle having a velocity v = v0 at t = 0 is decelerated at the rate a = v ,

where is a positive constant

A) The particle comes to rest at t = 0v2

B) The particle will come to rest at infinity

C) The distance traveled by the particle is

230

v2

D) The distance travelled by the particle is

230

v

3

2

25. Two particles P and Q move in a straight line AB towards each other. P startsfrom A with velocity u1 and an acceleration a1. Q starts from B with velocityu2 and acceleration a2. They pass each other at the midpoint of AB and arrive

at the other ends of AB with equal velocities


47/60

A) They meet at midpoint at time t =)aa(

)uu(2

21

12

B) The length of path specified i.e. AB is l =2

21

122112

)aa(

)uaua()uu(4

C) They reach the other ends of AB with equal velocities if (u2 + u1) (a1- a2) = 8(a1u2 - a2u1)D) They reach the other ends of AB with equal velocities if (u2 - u1) (a1 + a2) = 8(a2u1 - a1u2)

26. A body is moving along a straight line. Its distance xt from a point on its pathat a time t after passing that point is given by xt = 8t2 - 3t3, where xt is in metreand t is in secondA) Average speed during the interval t = 0 s to t = 4 s is 20.2 l ms-1B) Average velocity during the interval t = 0 s to t= 4 s is -16 ms-1

C) The body starts from rest and at t = 916 s it reverses its direction of motion

at xt = 8.43 m from the start.D) It has an acceleration of - 56 ms-2 at t = 4 s

27. Two second after projection, a projectile is traveling in a direction inclined at300 to the horizon. After one more second it is traveling horizontally. ThenA) the velocity of projection is 20 ms-1 B) the velocity of projection is 20

3 ms-1

C) the angle of projection is 300 with verticalD) the angle of projection is 300 with horizon

28. A shot is fired with a velocity u at an angle ( + ) with the horizon from thefoot of an incline plane of angle through the point of projection. If it hits theplane horizontally then

A) tan =+

2tan21

tanB) tan = 2 tan C) tan =

+

2tan21

tan2D) tan =

+

2sin1

cossin

29. A particle is projected with a velocity 2 hg so that it just clears two walls of

equal height h at horizontal separation 2h from each other. Then the

A) angle of projection is 300

with vertical B) angle of projection is 300

withhorizon

C) time of passing between the walls isg

h2

D) time of passing between the walls is 2g

h

30. A ball starts falling freely from a height h from a point on the

inclined plane forming an angle with the horizontal as shown.After collision with the incline it rebounds elastically off the inclinedplane. Then

A) it again strikes the incline at t =g

h8 after it strikes the incline at A


48/60

B) it again strikes the incline at t =g

h2after it strikes the incline at A

C) it again strikes the incline at a distance 4h sin from A along the incline

D) it again strikes the incline at a distance 8h sin from A along the incline31. Two particles projected from the same point with same speed u at angles of

projection and strike the horizontal ground at the same point. If h1 and h2are the maximum heights attained by projectiles, R be the range for both andt1 and t2 be their time of flights respectively then

A) + =2

B) R = 4 21hh C)

2

1

t

t= tan D) tan =

2

1

h

h

32. Two shells are fired from a cannon successively with speed u each at angles of

projection and respectively. If the time interval between the firing of shellsis t and they collide in mid air after a time T from the firing of the first shell.Then

A) T cos = (T - t) cos B) >

C) (T - t) cos = t cos

D) ( u sin ) T -2

1gT2 = ( u sin ) (T - t) -

2

1g(T - t)2

33. Two guns situated at the top of a hill of height 10 m, fire one shot each with

the same speed of 5 3 ms-1 at some interval of time. One gun fires

horizontally and other fires upwards at an angle of 600 with the horizontal.

The shots collide in mid air at the point P. Taking the origin of the coordinatesystem at the foot of the hill right below the muzzle, trajectories in x - y planeand g = 10 ms-2 thenA) the first shell reaches the point P at t1 = 1 s from the startB) the second shell reaches the point P at t2 = 2 s from the startC) the first shell is fired 1 s after the firing of the second shell

D) they collide at P whose coordinates are given by (5 3 , 5) m

34. A radar observer on the ground is watching an approaching projectile. At acertain instant he has the following information:i) The projectile has reached the maximum altitude and is moving with ahorizontal velocity v;ii) The straight line distance of the observer to the projectile is l;

iii) The line of sight to the projectile is at an angle above the horizontalAssuming earth to be flat and the observer lying in the plane of the projectile'strajectory then,A) the distance between the observer and the point of impact of the projectileis

D =g

sinv2

- l cos


49/60

B) the distance between the observer and the point of impact of the projectile is

D = vg

sin2 l- l cos

C) the projectile will pass over the observer's head for l

study material kinematics for aieee

Documents